If `ABCDEF` is a regular hexagon with `vec(AB) = veca` and `vec(BC)= vecb,` then `vec(CE)` equals

Options

A. `vec(b) - vec(a)`
B. `-vec(b)`
C. `vec(b)-2vec(a)`
D. none of these

Explanation

To solve the problem, we need to find the vector \(\vec{CE}\) in a regular hexagon \(ABCDEF\) where \(\vec{AB} = \vec{a}\) and \(\vec{BC} = \vec{b}\).

### Step-by-Step Solution:

1. **Understanding the Hexagon**:
- Draw a regular hexagon and label the vertices as \(A, B, C, D, E, F\).
- The sides of the hexagon are equal, and the internal angles are \(120^\circ\).

2. **Vectors Given**:
- We know that \(\vec{AB} = \vec{a}\) and \(\vec{BC} = \vec{b}\).

3. **Finding \(\vec{AC}\)**:
- By the triangle law of vector addition, we can find \(\vec{AC}\):
\[ \vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b} \]

4. **Finding \(\vec{AD}\)**:
- Since \(AD\) is parallel to \(BC\) and twice its length, we have:
\[ \vec{AD} = 2\vec{b} \]

5. **Finding \(\vec{CD}\)**:
- Using the relationship between the vectors, we can express \(\vec{CD}\):
\[ \vec{AD} = \vec{AC} + \vec{CD} \]
- Substituting the known values:
\[ 2\vec{b} = (\vec{a} + \vec{b}) + \vec{CD} \]
- Rearranging gives us:
\[ \vec{CD} = 2\vec{b} - (\vec{a} + \vec{b}) = \vec{b} - \vec{a} \]

6. **Finding \(\vec{CE}\)**:
- Now, we can find \(\vec{CE}\) using the relationship:
\[ \vec{CE} = \vec{CD} + \vec{DE} \]
- Since \(\vec{DE} = -\vec{a}\) (as \(DE\) is opposite to \(AB\)):
\[ \vec{CE} = (\vec{b} - \vec{a}) + (-\vec{a}) = \vec{b} - 2\vec{a} \]

### Final Answer:
\[ \vec{CE} = \vec{b} - 2\vec{a} \]